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Sagot :
Let's approach this problem step-by-step.
### Part (a): Finding [tex]\(\alpha^2 + \beta^2\)[/tex] in terms of [tex]\(k\)[/tex]
For the quadratic equation [tex]\(f(x) = x^2 + (k-3)x + 4 = 0\)[/tex], the roots are [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex].
Using Vieta's formulas:
- The sum of the roots [tex]\(\alpha + \beta = -(k-3)\)[/tex]
- The product of the roots [tex]\(\alpha \beta = 4\)[/tex]
We want to find [tex]\(\alpha^2 + \beta^2\)[/tex]. Using the identity:
[tex]\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \][/tex]
Substitute the known values:
[tex]\[ \alpha + \beta = -(k-3) \implies (\alpha + \beta)^2 = (-(k-3))^2 = (k-3)^2 \][/tex]
[tex]\[ \alpha \beta = 4 \][/tex]
So,
[tex]\[ \alpha^2 + \beta^2 = (k-3)^2 - 2 \cdot 4 = (k-3)^2 - 8 \][/tex]
### Part (b): Forming a quadratic equation with roots [tex]\(\frac{1}{\alpha^2}\)[/tex] and [tex]\(\frac{1}{\beta^2}\)[/tex]
Given [tex]\( \alpha \)[/tex] and [tex]\( \beta \)[/tex] are the roots of the original equation, we need to form a quadratic equation whose roots are [tex]\(\frac{1}{\alpha^2}\)[/tex] and [tex]\(\frac{1}{\beta^2}\)[/tex].
First, consider the sum and product of the new roots:
- Sum of the new roots: [tex]\(\frac{1}{\alpha^2} + \frac{1}{\beta^2}\)[/tex]
- Product of the new roots: [tex]\(\frac{1}{\alpha^2} \cdot \frac{1}{\beta^2}\)[/tex]
The product of the new roots is straightforward:
[tex]\[ \frac{1}{\alpha^2} \cdot \frac{1}{\beta^2} = \frac{1}{\alpha^2 \beta^2} = \frac{1}{(4)^2} = \frac{1}{16} \][/tex]
For the sum of the new roots, use:
[tex]\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{\alpha^2 \beta^2} = \frac{\alpha^2 + \beta^2}{(4)^2} = \frac{\alpha^2 + \beta^2}{16} \][/tex]
From part (a), we have:
[tex]\[ \alpha^2 + \beta^2 = (k-3)^2 - 8 \][/tex]
Therefore,
[tex]\[ \sum = \frac{(k-3)^2 - 8}{16} \][/tex]
The quadratic equation with roots [tex]\(\frac{1}{\alpha^2}\)[/tex] and [tex]\(\frac{1}{\beta^2}\)[/tex] is:
[tex]\[ x^2 - \left(\frac{(k-3)^2 - 8}{16}\right)x + \frac{1}{16} = 0 \][/tex]
Multiply through by 16 to get integer coefficients:
[tex]\[ 16x^2 - ((k-3)^2 - 8)x + 1 = 0 \][/tex]
### Part (c): Finding the possible values of [tex]\(k\)[/tex]
Given the condition:
[tex]\[ 4(\alpha^2 + \beta^2) = 7(\alpha \beta)^2 \][/tex]
We substitute the known values:
[tex]\[ 4((k-3)^2 - 8) = 7 \cdot 4^2 \][/tex]
[tex]\[ 4((k-3)^2 - 8) = 7 \cdot 16 \][/tex]
[tex]\[ 4(k-3)^2 - 32 = 112 \][/tex]
[tex]\[ 4(k-3)^2 = 144 \][/tex]
[tex]\[ (k-3)^2 = 36 \][/tex]
[tex]\[ k-3 = \pm 6 \][/tex]
[tex]\[ k = 3 + 6 \quad \text{or} \quad k = 3 - 6 \][/tex]
[tex]\[ k = 9 \quad \text{or} \quad k = -3 \][/tex]
Thus, the possible values of [tex]\(k\)[/tex] are [tex]\(9\)[/tex] and [tex]\(-3\)[/tex].
### Part (a): Finding [tex]\(\alpha^2 + \beta^2\)[/tex] in terms of [tex]\(k\)[/tex]
For the quadratic equation [tex]\(f(x) = x^2 + (k-3)x + 4 = 0\)[/tex], the roots are [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex].
Using Vieta's formulas:
- The sum of the roots [tex]\(\alpha + \beta = -(k-3)\)[/tex]
- The product of the roots [tex]\(\alpha \beta = 4\)[/tex]
We want to find [tex]\(\alpha^2 + \beta^2\)[/tex]. Using the identity:
[tex]\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \][/tex]
Substitute the known values:
[tex]\[ \alpha + \beta = -(k-3) \implies (\alpha + \beta)^2 = (-(k-3))^2 = (k-3)^2 \][/tex]
[tex]\[ \alpha \beta = 4 \][/tex]
So,
[tex]\[ \alpha^2 + \beta^2 = (k-3)^2 - 2 \cdot 4 = (k-3)^2 - 8 \][/tex]
### Part (b): Forming a quadratic equation with roots [tex]\(\frac{1}{\alpha^2}\)[/tex] and [tex]\(\frac{1}{\beta^2}\)[/tex]
Given [tex]\( \alpha \)[/tex] and [tex]\( \beta \)[/tex] are the roots of the original equation, we need to form a quadratic equation whose roots are [tex]\(\frac{1}{\alpha^2}\)[/tex] and [tex]\(\frac{1}{\beta^2}\)[/tex].
First, consider the sum and product of the new roots:
- Sum of the new roots: [tex]\(\frac{1}{\alpha^2} + \frac{1}{\beta^2}\)[/tex]
- Product of the new roots: [tex]\(\frac{1}{\alpha^2} \cdot \frac{1}{\beta^2}\)[/tex]
The product of the new roots is straightforward:
[tex]\[ \frac{1}{\alpha^2} \cdot \frac{1}{\beta^2} = \frac{1}{\alpha^2 \beta^2} = \frac{1}{(4)^2} = \frac{1}{16} \][/tex]
For the sum of the new roots, use:
[tex]\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{\alpha^2 \beta^2} = \frac{\alpha^2 + \beta^2}{(4)^2} = \frac{\alpha^2 + \beta^2}{16} \][/tex]
From part (a), we have:
[tex]\[ \alpha^2 + \beta^2 = (k-3)^2 - 8 \][/tex]
Therefore,
[tex]\[ \sum = \frac{(k-3)^2 - 8}{16} \][/tex]
The quadratic equation with roots [tex]\(\frac{1}{\alpha^2}\)[/tex] and [tex]\(\frac{1}{\beta^2}\)[/tex] is:
[tex]\[ x^2 - \left(\frac{(k-3)^2 - 8}{16}\right)x + \frac{1}{16} = 0 \][/tex]
Multiply through by 16 to get integer coefficients:
[tex]\[ 16x^2 - ((k-3)^2 - 8)x + 1 = 0 \][/tex]
### Part (c): Finding the possible values of [tex]\(k\)[/tex]
Given the condition:
[tex]\[ 4(\alpha^2 + \beta^2) = 7(\alpha \beta)^2 \][/tex]
We substitute the known values:
[tex]\[ 4((k-3)^2 - 8) = 7 \cdot 4^2 \][/tex]
[tex]\[ 4((k-3)^2 - 8) = 7 \cdot 16 \][/tex]
[tex]\[ 4(k-3)^2 - 32 = 112 \][/tex]
[tex]\[ 4(k-3)^2 = 144 \][/tex]
[tex]\[ (k-3)^2 = 36 \][/tex]
[tex]\[ k-3 = \pm 6 \][/tex]
[tex]\[ k = 3 + 6 \quad \text{or} \quad k = 3 - 6 \][/tex]
[tex]\[ k = 9 \quad \text{or} \quad k = -3 \][/tex]
Thus, the possible values of [tex]\(k\)[/tex] are [tex]\(9\)[/tex] and [tex]\(-3\)[/tex].
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